Question

How do you find the definite integral of `xsqrt(x - 40)dx` from `[40, 41]`?

Answer 1

`int_40^41xsqrt(x-40)color(white).dx=406/15`

Explanation:

`I=int_40^41xsqrt(x-40)color(white).dx`

We should apply the substitution `u=x-40`. This implies that `x=u+40` and that `du=dx`.

When doing the substitution, plug the current bounds of `x=40` and `x=41` into `u=x-40`. These give bounds of `x=40=>u=40-40=0` and `x=41=>u=41-40=1`. Thus:

`I=int_0^1(u+40)sqrtucolor(white).du`

Expanding this:

`I=int_0^1(u^1+40)u^(1/2)color(white).du`

`I=int_0^1(u^(3/2)+40u^(1/2))color(white).du`

Integrating these using `intu^ncolor(white).du=u^(n+1)/(n+1)+C`, and applying the bounds:

`I=[u^(5/2)/(5/2)+40(u^(3/2)/(3/2))]_0^1`

`I=[2/5u^(5/2)+80/3u^(3/2)]_0^1`

`I=[2/5(1)^(5/2)+80/3(1)^(3/2)]-[2/5(0)+80/3(0)]`

`I=2/5+80/3`

`I=406/15`